How to print the number of seconds since unix epoch in solaris shell

Question

Is there a way to use unix date to print the number of seconds since epoch?

I'm open to using other standard shell commands if there is a way

(I'm using Solaris, so there isn't date "+"%s")

Thanks

Is perl (perl -e 'print localtime' ) not an acceptable answer? I can't think of how to do it with the date command. — MJB, Apr 07 '10 at 12:56
@MJB rather than 'print localtime', the comparable perl expression to "date +%s" would be "perl -e 'print time'" — michael, Jul 13 '12 at 07:45
probably should be considered a duplicate of http://stackoverflow.com/questions/2445198/get-seconds-since-epoch-in-any-posix-compliant-shell (but there are many duplicates of this question) — michael, Jul 13 '12 at 07:46

score 6 · Answer 1 · answered Apr 07 '10 at 12:57

6

perl -e 'print time'

answered Apr 07 '10 at 12:57

Marcelo Cantos

score 3 · Accepted Answer · answered Apr 07 '10 at 12:57

3

Well, since this is a programming site, I would just compile something like:

#include <stdio.h>
#include <time.h>
int main (void) {
    printf ("%d\n",time(0));
    return 0;
}

:-)

answered Apr 07 '10 at 12:57

paxdiablo

score 0 · Answer 3 · answered Jul 08 '14 at 10:01

0

Try this

/usr/bin/truss /usr/bin/date 2>&1 | grep ^time | awk '{print $3}'

answered Jul 08 '14 at 10:01

Abhijeet Kasurde

score 0 · Answer 4 · answered Apr 07 '10 at 13:15

0

you can try nawk,(but try to double confirm with the perl commands and see if they give the same results)

nawk "BEGIN{print srand}"

answered Apr 07 '10 at 13:15

ghostdog74

4 Answers4